Denote by \(\lbrace _i\rbrace _{i=1}^m\)  the upper central series of \({u}{}\) .
By Zariski density of the \(k\) -points, the collection \(\lbrace Z_i\rbrace _{i=1}^m\)  is the upper central series of \({u}{G}\) .
Implication (1) to (2) is straightforward. Indeed, assume that \(G\)  is sealed since \(G\)  is Polish and any quotient group is sealed and in particular minimal. In particular, for any unipotent normal \(k\) -algebraic subgroup \(\le \)  the group \(G/N=/(k)\)  (see Proposition REF ) is minimal and by Lemma REF , its center is compact.
Assume (2), observe that \(L=G/{u}{G}\)  hence its center is compact. We will show that the action of \(L\)  on \({u}{G}\)  is with no non-trivial fixed points. Assume by contradiction that there exists a non-trivial element \(u\in {u}{G}\)  such that for any \(l\in L\) , \([l,u]=e\) . Let \(i\)  be the minimal index such that \(u\in Z_i\setminus Z_{i-1}\) . Observe that \(_{i-1}\)  is a normal unipotent \(k\) -algebraic group hence by assumption \(Z(G/Z_{i-1})\)  is compact. But then \(u Z_{i-1} \in Z(G/Z_{i-1}) \)  which is a contradiction since compact algebraic groups have no unipotent elements. Hence the action of \(L\)  on \({u}{G}\)  is with no non-trivial fixed points and (3) holds.
By the Lie-correspondence for unipotent algebraic groups in characteristic zero [1]}, the logarithmic mapping \({u}{}\rightarrow ({u}{}\)  is an \(\)  equivariant bijection.
Hence a non-trivial fixed point in the unipotent radical will be mapped to a non-trivial invariant vector in the Lie-Algebra, and vise versa, therefore (3) and (4) are equivalent.
We will prove that condition (3) implies (1) by induction on \(n\) , the nilpotency degree of the unipotent radical \({u}{}\) .
If \(n=0\) , that is \({u}{}=\lbrace e\rbrace \) , then the group \(\)  is reductive and \(G\)  has compact center, it follows by Proposition REF  that \(G\)  is sealed.
If \(n=1\) , that is \({u}{}\)  is abelian, it follows by Proposition REF  that \(G\)  is sealed.
Assume that the nilpotency degree of \({u}{}\)  is \(n>1\) .
Set \(_1 = \ltimes _1\) , and \(^1 = \ltimes {u}{} / _1\) . We will show that condition (3) holds for the \(k\) -groups \(_1\)  and \(^1\) .
Clearly \(Z(L)\)  is compact, and as \(_1\le {u}{}\) , the action of \(\)  on \(_1\)  is with non-trivial fixed point.
We are left to show that the \(\) -action on \({u}{} / _1\)  is with no non-trivial fixed points.
Assume by contradiction that there exists a non-trivial fixed point in \({u}{} / _1\) , i.e. there exists an element \(u\notin _1\)  such that for any \(l\in \) , \([l,u]\in _1\) .
The commutation map
\(\mathbf {c}_u \colon \rightarrow _1 \ \ \mathbf {c}_u(l)=[l,u]\) 
satisfies the following co-cycle relation
\(c_u(xy)=xyuy^{-1}x^{-1}u^{-1} = x(yuy^{-1}u^{-1})x^{-1} (x ux^{-1}u^{-1})= c_u(x)+x.c_u(y)\) 
with respect to the conjugation action of \(\)  on \(_1\) .
By [1]} the first co-homology of \(\)  is trivial, therefore there exist \(z\in _1\)  such that \( \mathbf {c}_u (l)=l.z-z=[l,z]\) . We get that for any \(l\in \)
\(e &= [l^{-1},u][l^{-1},z]^{-1} \\&= [l^{-1},u][z,l^{-1}] \\&= l^{-1}z^{-1} uz l u^{-1}z l^{-1} z^{-1}l\\&= l^{-1}z^{-1} [uz, l] z^{-1} l\\&= l^{-1} [uz, l] l\\\) 

hence for any \(l\in \) , \([uz,l]=e\)  in contradiction to \({u}{}\)  having no non-trivial fixed point.
This shows that the \(\) -action on \({u}{} / _1\)  is indeed with no non-trivial fixed points.
Since the nilpotency degrees of \({u}{_1}\)  and of \({u}{^1}\)  are strictly smaller then \(n\) , we get by the induction hypothesis that \(G_1\)  and \(G^1\)  are sealed.
Finally, let \(f\colon G\rightarrow H\) , be a continuous homomorphism into a topological group \(H\)  with a dense image.
The normal subgroup \(G_1\lhd G\)  is sealed, hence \(f(G_1)\)  is a normal closed subgroup of \(H\) , making \(H/f(G_1)\)  a topological group.
Since \(Z_1\le G_1\)  the map \(f\)  factors through \(f_{1}\colon G/(Z_1) \rightarrow H/f(G_1)\) .
Observe that \(G/Z_1\cong G^1\) . Consider the following commutative diagram
\(\begin{tikzcd}G ["f"][] & H []\\G^1["f_{1}"] & H/f(G_1)\end{tikzcd}\) 
The subgroup \(f_{1}(G^1)\)  is a closed subgroup in \(H/f(G_1)\)  and therefore its preimage in \(H\) , which is equal to \(f(G)\) , is closed. This means that \(G\)  is sealed, and the proof is complete.
